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Q.

Considering 1 g of water of volume $1 c m^{3}$ becomes $1681 c m^{3}$ of steam when boiled at 1 atm pressure. What is the external work done? ( $1 a t m = 10^{5} P a$ )

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a

268 J

b

168 J

c

468 J

d

368 J

answer is A.

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Detailed Solution

$W = p d v$

$= 10^{5} (1681 - 1) \times 10^{- 6}$

= 168 J

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